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Very recently, I learned a useful procedure due to J. P. Boyd (see also this and this) that involves expanding a function as a Chebyshev polynomial series, forming the so-called "colleague matrix", and then finding the eigenvalues of this matrix, which are often good approximations to the roots of the original function. I shall present how to do a barebones ...

The title of your question states that you will be satisfied with an instance of {p, n} that satisfies Sum[p^k/k!, {k, 0, n}] > 0. Here is a function to find any number of such instances:
instances[m_Integer?Positive] :=
Block[{p, n, k},
FindInstance[Sum[p^k/k!, {k, 0, n}] > 0 && p ∈ Reals && n ∈ Integers, {n, p}, m]]
And ...

Here is a partial answer. I believe for Method -> {opts}, opts may be any of the following, and they will have whatever effect they have:
Internal`InequalitySolverOptions[]
Internal`ReduceOptions[]
Internal`NSolveOptions[]
(*
{"ARSDecision" -> False, "BrownProjection" -> True, "CAD" -> True,
"CADAlgebraicCoefficients" -> True, ...

How are you calculating the velocity around the cylinder I am using
Plot[Norm[f[20 + 5 Cos[\[Theta]], 20 + 5 Sin[ \[Theta]]]], {\[Theta],
0, 2 \[Pi]}]
to give
Which I think is correct.
For the pressure we need the correct form for Bernoulli.
Where you take the values of pressure at infinity as 0 but ignore the velocity at infinity. I am also ...

I know nothing about the range of the data for your problem. Assuming that the values for Rho, Mu and z are real numbers you can gain insight into your problem by combining bbgodfrey's comment with a plot using Manipulate.
For example, if Rho and Mu are known parameters you can see how the solution varies as you change the value of z.
Manipulate[
...

Can I create a function which efficiently returns a list of the small positive roots of $\varepsilon \lambda=\cot\lambda$?
Here is a routine I wrote quite a while back, when I was doing some research related to the square well potential. The procedure does the computation in two stages: an initial approximation computed through the Delves-Lyness contour ...

The issue is that you're asking for a general solution for y only whereas your particular set of equations has a solution only for one specific value of x.
You should have called Solve as:
Solve[f[x, y] == g[x, y], {x, y}]
During evaluation of Solve::ratnz: Solve was unable to solve the system with inexact coefficients. The answer was obtained by ...

FindInstance, as described by Mr.Wizard, gives solutions for a given perimeter, which may then be tested for uniqueness. However, Project Euler challenges you to consider and learn about other methods which solve the problem, usually faster and without a mysterious "black box".
For example, this problem concerns right-angled triangles with integer sides, ...

This is a hack to create a function which interoperates at least on some level with FindRoot. Still, it's very much only a starting point. Primary goal of regularize is to attempt keeping values below and above sought root not jumping on the other side.
ClearAll@regularize;
regularize[f_Function, rootval_, minsigmas_, minsamples_Integer,
...

You can use bezier from my answer to How to know form of plotted Bezier function to get the formulas:
bezier[pts_List] :=
With[{n = Length[pts] - 1},
Evaluate@ Sum[Binomial[n, i] (1 - #)^(n - i) #^i pts[[i + 1]], {i, 0, n}] &]
Following the explanation in the linked answer, we can solve the equation bezier[pp][t] == {x, y} for y in terms of x as ...

One way to deal with the singularities (that lead to dimensional components of the solution space) when x, y and/or z are zero is through a change of variables that lets excessive zeros be factored out.
changevar = {y -> x s, z -> x s t};
Short /@ Factor[foc /. changevar]
We can use FactorList to get at the irreducible factors and delete constants ...

This is more an extended comment, although it does answer the question. Using Belisarius' interpretation of the equations, one might think that
Solve[{eq1, eq4}, {A1, m1}]
could yield a solution directly, but it returns unevaluated. However, Solve can yield a solution, if provided some hand-holding.
solve[{e1_, e2_}, {a_, b_}] := Module[{t1, t2, t3}, ...

You're in luck if you're only interested in a domain where x = h/(2 s0) is approximately 1/2. That domain is contained by the domain of InverseFunction applied to Cos[φ0]/φ0:
ϕ = InverseFunction[Cos[#]/# &]
InverseFunction can only work on a domain on which the function is one-to-one. In this case, it's from zero to the first minimum around ϕ0 -> ...

Alexei Boulbitch's answer is quite nice, but here are another couple of useful techniques for situations like this. First, if all you need is a graph of the function, you can use ParametricPlot to get that. You want $\phi(x)$, where $\cos \phi/\phi = x$; so you can use $\phi$ as a parameter and plot the curve $(\cos \phi/\phi, \phi)$:
...

Looking at the plot:
Manipulate[Plot[Cos[x]/x - y, {x, 0, 2 \[Pi]}], {y, 0, 1}]
that is, here:
and varying the parameter y=h0/(2s0), one finds that the first root is close to x==1. Let us look for this root only, than it is easy to find a list of values with the structure {y,x0}, where y is the value of the parameter and x0 is the root:
lst = ...

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